Question: The differentiable functions $x$ and $y$ are related by the following equation: $3y=\cos(x)$ Also, $\dfrac{dy}{dt}=5$. Find $\dfrac{dx}{dt}$ when $x=\dfrac{\pi}{2}$.
Solution: Let's start by differentiating the equation $3y=\cos(x)$ with respect to $t$. $\begin{aligned} 3y&=\cos(x) \\\\ 3\cdot\dfrac{dy}{dt}&=-\sin(x)\cdot\dfrac{dx}{dt} \end{aligned}$ We are given that $\dfrac{dy}{dt}=5$, and we want to find $\dfrac{dx}{dt}$ when $x=\dfrac{\pi}{2}$. Let's plug ${x=\dfrac{\pi}{2}}$ and ${\dfrac{dy}{dt}=5}$ into the equation we obtained: $\begin{aligned} 3\cdot{\dfrac{dy}{dt}}&=-\sin({x})\cdot\dfrac{dx}{dt} \\\\ 3\cdot{5}&=-\sin\left({\dfrac{\pi}{2}}\right)\cdot\dfrac{dx}{dt} \\\\ 15&=-1\cdot \dfrac{dx}{dt} \\\\ -15&=\dfrac{dx}{dt} \end{aligned}$ In conclusion, when $x=\dfrac{\pi}{2}$, the value of $\dfrac{dx}{dt}$ is $-15$.